Quick question on performance

[vladibo]

What is more efficient:

A. if (!setWidth) { DoWork() }

or

B. if (typeof setWidth == 'undefined') { DoWork() }

 

[trollacious]

if (!setWidth)

is NOT a check to see if setWidth is defined or not. If setWidth is zero, it will pass the conditional test you have. The 2 examples you have will not provide the same results, which makes your question moot.

Lee

 

[BabyJeffy]

Besides which... running that kind of test produces javascript errors. I'd put it within an try/catch but that defeats the purpose of any testing. And yeah... the first one fails if the object is zero.



Jeff

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[vladibo]

In my case the object is always of type Function so my question is actually: Does anybody know what operations are performed on a lower level on (!Variable)

My code looks like:

if (!setWidth) { setWidth = new Function("return something;") }

 

[BillyRayPreachersSon]

Does anybody know what operations are performed on a lower level on (!Variable)

I have no idea what you mean by this. What is "a lower level"? Can you explain more?

Dan



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[BabyJeffy]

object is always of type Function

So what is the point of doing the following?

 if (typeof setWidth == 'undefined') { DoWork() }

Jeff

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[LuckyKoen]

I usually use 2 dates to test the processing time of a certain block of code. Define one Date before and one after the code and substract the 2nd from the first. An easy way to see how long it took to process the code.

 

[vladibo]

I mean if on the C++ level the operations for (!variabale) are

1. Retrieve the variable type as string
2. Check if is null
3. Compare this string with 'undefined'
4. Check if is 0

Then if (typeof setWidth == 'undefined') { DoWork() } might be even more efficient

But if it is

1. Check how many bit are associated with the space allocated for this variable

then (!setWidth) is better

 

[vladibo]

This is a way to check if a function is defined and if not to define it

 

[BillyRayPreachersSon]

This is a way to check if a function is defined and if not to define it

Erm.. no it is not. You were running a function called "DoWork". You are never defining a function anywhere depending on the outcome of the test.

Dan



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